Suppose that \(\mathbf {c}=\mathbf {0}\) . Recalling (REF ) and the notation \(r=q/Q\) , it is clear that
\(q^k\frac{\partial ^kI_{q,\mathbf {a}}(\mathbf {0})}{\partial q^k} = r^{k}P_1\cdots P_4\int _{\mathbb {R}^4}w(\mathbf {x})\frac{\partial ^k h(r,G(\mathbf {x}))}{\partial r^k} \mathrm {d}\mathbf {x}.\) 
From [1]} with \(m=k, n=0\)  and \(N=2\) , we have
\(\frac{\partial ^k h(r,G(\mathbf {x}))}{\partial r^k}\ll r^{-1-k}\left(r^2+ \min \left\lbrace 1,\frac{r^2}{|G(\mathbf {x})|^2}\right\rbrace \right),\) 
and so
\((P_1\cdots P_4)^{-1}\frac{\partial ^kI_{q,\mathbf {a}}(\mathbf {c})}{\partial q^k} \ll r + r^{-1}\int _{\begin{array}{c}\mathbf {x}\in \operatorname{supp}w\\ |G(\mathbf {x})|\leqslant r\end{array}}1\mathrm {d}\mathbf {x}+ r\int _{\begin{array}{c}\mathbf {x}\in \operatorname{supp}w\\|G(\mathbf {x})|>r\end{array}}\frac{1}{|G(\mathbf {x})|^2}\mathrm {d}\mathbf {x}. \) 
We need to show that the right hand side of (REF ) is \(O(1)\) . We can assume that the first term \(r\)  is \(O(1)\) , since \(h(r,G(\mathbf {x}))=0\)  when \(r>\max (1,|G(\mathbf {x})|)\) , and \(G(\mathbf {x}) \ll 1\)  for \(x \in \operatorname{supp}w\) , and so \(I_{q,\mathbf {a}}(\mathbf {0})=0\)  unless \(r \ll 1\) . In order to estimate the integrals in (REF ), we consider for \(z>0\)  the Lebesgue measure \(m(z)\)  of the set
\(S(z)=\lbrace \mathbf {x}\in \operatorname{supp}w:|G(\mathbf {x})|\leqslant z\rbrace .\) 
We can find distinct indices \(i,j\)  such that \(\epsilon _i = \epsilon _j\) . Without loss of generality we may assume \(i=1, j=2\) , and in addition that \(\epsilon _1 = \epsilon _2 = 1\) . For a fixed choice of \(x_3, x_4\) , we define \(c = \epsilon _3x_3^2 + \epsilon _4x_4^2\) , so that if \(\mathbf {x}\in S(z)\)  then \(x_1^2+x_2^2 \in [-c-z,-c+z]\) . The measure of the set of pairs \(x_1,x_2\)  satisfying this condition is \(O(z)\) . For \(\mathbf {x}\in \operatorname{supp}w\) , we require \(|x_3|, |x_4| \ll 1\) , and hence \(m(z) = O(z)\) . From this we see that the first integral in (REF ) is \(O(r)\) , and by breaking into dyadic intervals, the second integral is \(O(r^{-1})\) . Therefore the right hand side of (REF ) is \(O(1)\) , as required.
When \(\mathbf {c}\ne \mathbf {0}\) , we first note that for \(k=0\) , we can reduce to the case above by applying the trivial estimate to the extra exponential factor \(e(-\mathbf {v}\cdot \mathbf {x})\)  appearing in the integral defining \(I_{q,\mathbf {a}}(\mathbf {c})\) . It remains only to deal with the case \(\mathbf {c}\ne \mathbf {0}, k=1\) . Define the vector \(\mathbf {v}\in \mathbb {R}^4\)  by \(v_i = q^{-1}P_ic_i\) . Similarly to (REF ), we obtain
\(&(P_1 \cdots P_4)^{-1}q\frac{\partial I_{q,\mathbf {a}}(\mathbf {c})}{\partial q} \ll r\int _{\mathbb {R}^4}w(\mathbf {x})\frac{\partial \lbrace h(r,G(\mathbf {x}))e(-\mathbf {v}\cdot \mathbf {x})\rbrace }{\partial r} \mathrm {d}\mathbf {x}\\& = r\int _{\mathbb {R}^4}w(\mathbf {x})\left(\frac{\partial h(r,G(\mathbf {x}))}{\partial r}e(-\mathbf {v}\cdot \mathbf {x}) + h(r,G(\mathbf {x}))\frac{-2\pi i \mathbf {v}\cdot \mathbf {x}}{r}e(-\mathbf {v}\cdot \mathbf {x})\right) \mathrm {d}\mathbf {x}.\) 

The first term can again be dealt with using the trivial estimate for \(e(-\mathbf {v}\cdot \mathbf {x})\)  and the argument for the case \(\mathbf {c}= \mathbf {0}\)  given above. In order to estimate the second term, as in [1]}, we apply the divergence theorem
\(\int _{\mathbb {R}^4}\nabla \cdot (w(\mathbf {x})h(r,G(\mathbf {x})e(-\mathbf {v}\cdot \mathbf {x})\mathbf {x})\mathrm {d}\mathbf {x}= 0\) 
in order to remove the additional factor of \(2\pi i \mathbf {v}\cdot \mathbf {x}\) . This yields
\(\begin{split}&\int _{\mathbb {R}^4}w(\mathbf {x})h(r,G(\mathbf {x}))e(-\mathbf {v}\cdot \mathbf {x})(2\pi i \mathbf {v}\cdot \mathbf {x})\mathrm {d}\mathbf {x}=\int _{\mathbb {R}^4}\widetilde{w}(\mathbf {x})h(r,G(\mathbf {x}))e(-\mathbf {v}\cdot \mathbf {x})\mathrm {d}\mathbf {x}\\+&\int _{\mathbb {R}^4}4w(\mathbf {x}) h(r,G(\mathbf {x}))e(-\mathbf {v}\cdot \mathbf {x})\mathrm {d}\mathbf {x}+\int _{\mathbb {R}^4}2w(\mathbf {x})G(\mathbf {x})e(-\mathbf {v}\cdot \mathbf {x})\frac{\partial h(r,G(\mathbf {x}))}{\partial G(\mathbf {x})}\mathrm {d}\mathbf {x},\end{split}\) 
where \(\widetilde{w}(\mathbf {x})= (\mathbf {x}\cdot \nabla )w(\mathbf {x})\) . The second integral on the right hand side of (REF ) can now be treated in the same way as the easier cases \(k=0\)  or \(\mathbf {c}= \mathbf {0}\)  using the trivial estimate for \(e(-\mathbf {v}\cdot \mathbf {x})\) . The third integral can similarly be estimated by noting that \(G(\mathbf {x})\ll 1\)  for \(\mathbf {x}\in \operatorname{supp}(w)\) , and applying [1]} with \(n=1,m=0,N=0\) . For the first integral, we note that \(\widetilde{w}(\mathbf {x})\)  is uniformly bounded by \(\eta ^{-1}\) . Hence the above arguments can be applied to the first integral as well, but with an extra factor of \(\eta ^{-1}\) .
